A216188 Number of unordered pairs of anagrammatic (positive) integers adding to n.

0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 3, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 3, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0

Offset: 1

Author

Christian Schulz, Mar 11 2013

Keywords

• nonn
• base

Comments

Two positive integers are here defined as "anagrammatic" if (in base 10) they have the same number of 0 digits, 1 digits, 2 digits, ..., 9 digits. Thus, 123 and 231 are anagrammatic, but not 301 and 013, as leading zeros are omitted.

Examples

			For n = 88, the a(88) = 4 pairs are {17,71}, {26,62}, {35,53}, and {44,44}. For n = 609, the a(609) = 1 pair is {237,372}.
		

Links

• Christian Schulz, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)

Programs

Maple

getDigit := (n,k) -> floor(n/10^k) mod 10; getMaxDigit := n -> floor(log10(n)) + 1; getDigitMultiset := n -> convert([seq(getDigit(n,k),k=0..getMaxDigit(n)-1)],multiset); isAnagram := (m,n) -> evalb(getDigitMultiset(m) = getDigitMultiset(n)); A216188 := n -> convert([seq(eval(isAnagram(k,n-k),[true=1,false=0]),k=1..floor(n/2))],`+`); seq(A216188(n),n=1..50)

Mathematica

IsAnagram[x_, y_, b_: 10] := Sort[Permutations[IntegerDigits[x, b]]] == Sort[Permutations[IntegerDigits[y, b]]]; FindAnagramSums[n_, b_: 10] := Select[Table[{k, n - k}, {k, 0, Floor[n/2]}], IsAnagram[#[[1]], #[[2]], b] &]; Table[Length[FindAnagramSums[n]], {n, 1, 200}]